• Feb 5th 2009, 10:42 PM
reshma
ABCD is a cyclic quadrilateral. Diagonals of ABCD (ie.,AC & BD) meet at right angles at P.PL is a perpendicular bisector to chord AB. If PL is extended to meet chord DC at M, show that PM bisects DC.
• Feb 6th 2009, 08:07 AM
red_dog
$AL=LB, \ PL\perp AB\Rightarrow\Delta APB$ isosceles $\Rightarrow\widehat{PAB}=\widehat{PBA}$

But $\widehat{PAB}=\widehat{PDC}, \ \widehat{PBA}=\widehat{PCD}\Rightarrow\widehat{PDC }=\widehat{PCD}\Rightarrow\Delta PCD$ isosceles.

We have $\widehat{ACD}=\widehat{CAB}\Rightarrow AB\parallel DC\Rightarrow PM\perp CD\Rightarrow DM=MC$
• Feb 9th 2009, 11:26 AM
fardeen_gen
1)angle APB and DPC are vertically opposite,hence equal
2)opposite sides are similar in ABCD because the diagonals are perpendicular to each other, hence have equal length
3)angle PLA and angle PMD are alternate angles, hence equal

Therefore, by ASA(angle-side-angle) postulate, triangles ABP and DCP are congruent.

Also,
PL=PM
Therefore, if PL bisects AB then, PM bisects DC.